First, another name for a proper topology on a function space is *splitting topology*, and other names for an admissible topology are *approximating topology* and *conjoining topology*. This may be of aid in further Google searches.

Here are some basic facts:

Any topology that is coarser (smaller) than a proper topology is also proper. Any topology that is finer (larger) than an admissible topology is also admissible. (The proofs are easy.)

Any proper topology on $C(X, Y)$ is coarser than any admissible topology on $C(X, Y)$. See Lemma 2.2 here. It follows that any proper topology is contained in the intersection of all admissible topologies.

The intersection of all admissible topologies is a proper topology, called the *natural topology* on $C(X, Y)$. (This is subsumed under Proposition 5.13 that is cited below, but see also here.)

It follows that the natural topology is the unique maximal topology among all proper topologies: a topology is proper iff it is coarser than the natural topology. The natural topology is in general not admissible; it is admissible iff $C(X, Y)$ is an exponential object (i.e., represents the functor $\hom_{\text{Top}}(- \times X, Y): \text{Top} \to \text{Set}$ in the usual sense of category theory: there is an isomorphism $\hom_{\text{Top}}(-, C(X, Y)) \cong \hom_{\text{Top}}(- \times X, Y)$, induced in Yoneda-wise fashion from a map $C(X, Y) \times X \to Y$ which is the evaluation map).

So a reasonable answer to the question is given by understanding a little more concretely the natural topology on $C(X, Y)$. It may be described in terms of filter convergence as follows: let us say that a filter $\Phi$ on $C(X, Y)$ *converges continuously* to $f \in C(X, Y)$ if, whenever a filter $\mathcal{G}$ on $X$ contains the filter of neighborhoods of $b \in X$, then the filter on $Y$ generated by the filter base $\{F(G): F \in \Phi, G \in \mathcal{G}\}$ contains the filter of neighborhoods of $f(b)$, where we put $F(G) := \bigcup \{f(G): f \in F\}$. As usual when we have a filter convergence relation, we go on to define $U \in C(X, Y)$ to be *open* with respect to continuous convergence if $U$ belongs to any filter $\Phi$ that converges to a point in $U$.

**Theorem** (see Proposition 5.13 here): For any spaces $X, Y$, the topology of continuous convergence on $C(X, Y)$ coincides with the natural topology on $C(X, Y)$.

In summary: a topology $T$ on a function space $C(X, Y)$ is proper iff every $T$-open is also an open of every topology for which the evaluation map is continuous; equivalently, is an open with respect to continuous convergence.